Question

Математический анализ.
Пределы.
Решите.

Answer 1

187

$\displaystyle\sf \lim_{x \to \infty} \frac{2x+3}{x+\sqrt[3]{x}}= \Big [\left { {{|:x} \atop {|:x}} \right. \Big ] =\lim_{x \to \infty} \frac{2+\frac{3}{x}}{1+\sqrt[3]{\frac{1}{x^2}}}=\Big [\frac{2}{1}\Big ]=2$

188

$\displaystyle \sf \lim_{x \to \infty} \frac{x^2}{x\sqrt{x}+10} =\lim_{x \to \infty} \frac{x^2}{x^{\frac{3}{2}}+10}=\Big [2>\frac{3}{2}\Big ] =\Big [ \left { {{|:x^2} \atop {|:x^2}} \right. \Big ]=\lim_{x \to \infty}\frac{1}{\frac{1}{\sqrt{x}}+\frac{10}{x^2}}= \\=\Big [ \frac{1}{0} \Big ] =\infty$

189

$\displaystyle \lim_{x \to \infty} \frac{\sqrt[3]{x^2+1} }{x+1}=\Big [ x^{\frac{2}{3}}<x^1\Big ] = \Big [\left { {{|:x} \atop {|:x}} \right. \Big ]= \lim_{x \to \infty} \frac{\sqrt[3]{\frac{1}{x}+\frac{1}{x^3}}}{1+\frac{1}{x}}=\Big[\frac{0}{1}\Big] = 0$